Q) Show that the line with the equation 2x - 3y + 26 = 0 is a tangent to the circle with the equation x^2 + y^2 - 4x + 6y - 104 = 0. If T is the point of contact of the tangent with the circle, find the equation of the normal to the circle at T. Find also the coordinates of the point on the circle which is diametrically opposite to T.
Q) Show that the line with the equation 2x - 3y + 26 = 0 is a tangent to the circle with the equation x^2 + y^2 - 4x + 6y - 104 = 0. If T is the point of contact of the tangent with the circle, find the equation of the normal to the circle at T. Find also the coordinates of the point on the circle which is diametrically opposite to T.
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For it to be a tangent there will only be 1 point at which the line and circle meet, so there will be only set of values to the simultaneous equation. Show this. If you want to find the normal then you know the equation of the point and the gradient of the tangent. As the product of the gradient of the tangent and normal is -1 (they are perpendicular) you can hence find the equation of the normal in the form y-y1=m(x-x1). The point which is diametrically opposite is the other point of intersection of the normal with the circle. You know that the normal is an extended diameter as it is perpendicular to the circle at the point on intersection.
Q) Show that the line with the equation 2x - 3y + 26 = 0 is a tangent to the circle with the equation x^2 + y^2 - 4x + 6y - 104 = 0. 3y=2x+26.
y=(2/3)(x+13) Substitution into the circle equation gives: x^2 + (4/9)(x^2+26x+169) - 4x + 4(x+13) - 104=0 9x^2 + 4(x^2+26x+169) - 36x + 36(x+13) - (104)(9)=0 x^2[13] + x[(4)(26)-36+36] + 4(169)+(36)(13)-(104)(9)=0 13x^2 + 104x +208=0 x=-104+-rt[10816-4(13)(208)] / 26 = -104+-0 / 26 = -4. There is only one point at which the circle and line touch and thus it is a tangent. As x=-4 then y=6 and so the coordinate is (-4,6). Your working was probably correct (if you found there was only one root) but I got rid of the fractions by multiplying the equation by 9 at one point.